Question

[Nguyen Duc Trung Thanh]

Cho A=1/2²+1/3²+1/4²+...+1/100²

CMR:A<3/4

Answer 1

[Đức Phạm]

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\)

\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)

\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{101}\)

\(\Leftrightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{99}{202}< \frac{3}{4}\)

\(\Rightarrow A< \frac{3}{4}\left(đpcm\right)\)

Answer 2

[Nguyễn Anh Quân]

A= 1/4 +1/3^2 +1/4^2 +.....+ 1/100^2

< 1/4 + 1/2.3 + 1/3.4 +.....+1/99.100

=1/4 + 1/2-1/3+1/3-1/4+......+1/99-1/100

=1/4 +1/2 - 1/100 < 1/4+1/2 = 3/4

=> ĐPCM